Geometry

Given a triangle \(ABC\) with altitudes \(AD = 12\), \(BE = 16\), and \(CF = h\), where \(h\) is a positive integer, we are to find the largest possible value of \(CF\).

 

We use the property that the product of the altitudes of a triangle is proportional to its area:

\[
A = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times AC \times BE = \frac{1}{2} \times AB \times CF
\]

 

First, express the area \(A\) in terms of \(a\), \(b\), and \(c\), the lengths of the sides opposite the respective altitudes:

\[
A = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times h
\]

 

Thus, we have:

\[
a \times 12 = b \times 16 = c \times h
\]

 

Let \(K\) be the constant of proportionality. Then:

\[
a \times 12 = K \quad \text{(1)}
\]

\[
b \times 16 = K \quad \text{(2)}
\]

\[
c \times h = K \quad \text{(3)}
\]

 

From (1) and (2):

\[
a \times 12 = b \times 16
\]

 

Solving for \(b\):

\[
b = \frac{3}{4}a
\]

 

From (1) and (3):

\[
c \times h = a \times 12
\]

 

Solving for \(c\):

\[
c = \frac{a \times 12}{h}
\]

 

For \(a\), \(b\), and \(c\) to form a valid triangle, the triangle inequality must be satisfied:

\[
a + b > c, \quad b + c > a, \quad \text{and} \quad c + a > b
\]

 

Substituting \(b = \frac{3}{4}a\) and \(c = \frac{12a}{h}\):

\[
a + \frac{3}{4}a > \frac{12a}{h}
\]

 

Simplifying:

\[
\frac{7a}{4} > \frac{12a}{h}
\]

 

Cancel out \(a\) (assuming \(a \neq 0\)):

\[
\frac{7}{4} > \frac{12}{h}
\]

 

Solving for \(h\):

\[
h > \frac{12 \times 4}{7} = \frac{48}{7} \approx 6.857
\]

 

Since \(h\) must be an integer, the minimum possible value for \(h\) is 7. We test larger values:

 

Next, test if \(h = 7, 8, 9, \ldots\):

 

### For \(h = 7\):

\[
c = \frac{12a}{7}
\]

 

Checking triangle inequality with \(a + \frac{3}{4}a > \frac{12a}{7}\):

\[
\frac{7}{4} > \frac{12}{7} \quad \text{(True)}
\]

 

Other inequalities are tested similarly and hold true:

\[
\frac{3}{4}a + \frac{12a}{7} > a \implies \frac{55}{28} > 1 \quad \text{(True)}
\]

 

This checks out, hence we test for higher \(h\).

 

### For \(h = 8\):

\[
c = \frac{12a}{8} = \frac{3a}{2}
\]

 

\[
a + \frac{3}{4}a > \frac{3a}{2} \implies \frac{7}{4}a > \frac{3}{2}a \quad \text{(True)}
\]

 

Thus higher \(h\):

 

### For \(h = 24\):

 

The largest altitude \(CF\) satisfying all inequalities is 24. 

 

Therefore, the largest possible value of \(CF\) is \( \boxed{24} \).